Prove the following:
$ \sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=0 $

To do:

We have to prove that \( \sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=0 \).

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

$cos\ (90^{\circ}- \theta) = sin\ \theta$

Therefore,

$\sin \theta \sin \left(90^{\circ}-\theta\right)-\cos \theta \cos \left(90^{\circ}-\theta\right)=\sin \theta \cos \theta-\cos \theta \sin \theta$

$=\sin \theta \cos \theta-\sin \theta \cos \theta$

$=0$

Hence proved.   

Tutorialspoint

Updated on: 10-Oct-2022

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